National Math Olympiad

The lines $DE$ and $BC$ are parallel, so $\angle DCB=\angle CDE$. The inscribed angles over the chord $EF$ in the cyclic quadrilateral $ADFE$ are equal, $\angle FDE=\angle FAE$. This implies $$\angle DCB=\angle CDE=\angle FDE=\angle FAE.$$ The lines $DE$ and $BC$ are parallel, so $\angle ABC=\angle ADE$. Since the points $A,D,E$ and $F$ are concyclic, we have $\angle ADE=\angle AFE$, and so $\angle DBC=\angle EFA$. The triangles $AFE$ and $CBD$ have two angles in common, $\angle AFE=\angle DBC$ and $\angle FAE=\angle DCB$, hence they are similar.